the integral i get is [tex]\frac{-ln(2s-\sqrt{A})}{4\sqrt{A}}+\frac{ln(2s+\sqrt{A})}{4\sqrt{A}}[/tex] by using integration by partial fractions or
[tex]\frac{1}{2}tan^{-1}\frac{2s}{\sqrt{a-4s^{2}}}[/tex]

The function isn't wrong, the velocity is given in terms of its position.
I'm thinking what will happen is the A will cancel out when i do a definite integral, but i cant ever get A by itself, it is always a co-efficient of s.

when you say differentiate velocity, i'm assuming you mean with respect to time.
When you differentiate the RHS is dissappears because it is independent of t in the first place.

When you differentiate the RHS is dissappears because it is independent of t in the first place.

Careful, just because there is no explicit time dependence, does not mean that there is no implicit time dependence. The displacement 's' will depend on time won't it?....You will need to use the chain rule on this one.